witbout actually finding zeroes of polynomial f(x)=3(x^2-1)+2x-5 , find sum and product of zeroes
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sum of zeros = - coefficient of x/coefficient of x^2= -2/3
product of zeros= constant term/ coefficient of x^2= -8/3
Hope this helps
product of zeros= constant term/ coefficient of x^2= -8/3
Hope this helps
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Answer:
sum of the zeroes =-b/a
=-2/3
product of the zeroes =c/a
=-8/3
Step-by-step explanation:
f(x)=3(x^2-1)+2x-5
= 3x^2-3+2x-5
= 3x^2+2x-8
a=3 ;b=2 ;c=-8
sum of the zeroes =-b/a
=-2/3
product of the zeroes =c/a
=-8/3
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